Consider the ODE
$$\frac{dy}{dt}=f(y,t), \quad y(0)=a,$$
where we may assume that $f$ is Lipschitz continuous and there is a unique solution $y(t)$ on $[0,2]$.
My questions is following:
Is $y(1)$ a monotone (more precisely, increasing) function of $a$ ?
I believe the claim holds true because the direction fields never cross, and thus if a solution curve $y_1(t)$ is above $y_2(t)$ at $t=0$, then it will always be above $y_2$. If you look at a simple example like logistic equation, you'll see that $y(1)$ is indeed increasing with respect to $y(0)$. The continuous dependence on initial data has been well-studied but I can not find any result or even discussion on the monotonicity.
The answer is yes, and this can be proved following your reasoning. Let's suppose, by contradiction, that there exists $b>a$ such that $y(1,b)<y(1,a)$, where $y(t,y_0)$ is the (unique) solution to $$ \frac{dy}{dt}=f(y,t), \quad y(0)=y_0. \tag{1} $$ Since $y(t,a)$ and $y(t,b)$ are continuous functions of $t$ in $[0,1]$, $y(0,b)>y(0,a)$, and $y(1,b)<y(1,a)$, it follows from Bolzano's theorem that there exists $\tau\in(0,1)$ such that $y(\tau,b)=y(\tau,a)=y_{\tau}$. This implies that both $y(t,a)$ and $y(t,b)$ are solutions to the initial value problem $$ \frac{dy}{dt}=f(y,t), \quad y(\tau)=y_{\tau}, \tag{2} $$ but this contradicts the condition of uniqueness of the solution in $[0,2]$. Therefore, our hypothesis that there exists $b>a$ such that $y(1,b)<y(1,a)$ is false, and we may conclude that $y(1,a)$ is a non decreasing function of $a$. (As a matter of fact, $y(1,a)$ is an increasing function of $a$, since $y(1,b)=y(1,a)$ for $b>a$ would also violate the condition of uniqueness of the solution.)